Simplify and expand the following expression: $ \dfrac{3}{2r + 12}+ \dfrac{3}{5r + 30}- \dfrac{4}{r^2 + 12r + 36} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{3}{2r + 12} = \dfrac{3}{2(r + 6)}$ We can factor a $5$ out of denominator in the second term: $ \dfrac{3}{5r + 30} = \dfrac{3}{5(r + 6)}$ We can factor the quadratic in the third term: $ \dfrac{4}{r^2 + 12r + 36} = \dfrac{4}{(r + 6)(r + 6)}$ Now we have: $ \dfrac{3}{2(r + 6)}+ \dfrac{3}{5(r + 6)}- \dfrac{4}{(r + 6)(r + 6)} $ The least common multiple of the denominators is: $ 10(r + 6)(r + 6)$ In order to get the first term over $10(r + 6)(r + 6)$ , multiply by $\dfrac{5(r + 6)}{5(r + 6)}$ $ \dfrac{3}{2(r + 6)} \times \dfrac{5(r + 6)}{5(r + 6)} = \dfrac{15(r + 6)}{10(r + 6)(r + 6)} $ In order to get the second term over $10(r + 6)(r + 6)$ , multiply by $\dfrac{2(r + 6)}{2(r + 6)}$ $ \dfrac{3}{5(r + 6)} \times \dfrac{2(r + 6)}{2(r + 6)} = \dfrac{6(r + 6)}{10(r + 6)(r + 6)} $ In order to get the third term over $10(r + 6)(r + 6)$ , multiply by $\dfrac{10}{10}$ $ \dfrac{4}{(r + 6)(r + 6)} \times \dfrac{10}{10} = \dfrac{40}{10(r + 6)(r + 6)} $ Now we have: $ \dfrac{15(r + 6)}{10(r + 6)(r + 6)} + \dfrac{6(r + 6)}{10(r + 6)(r + 6)} - \dfrac{40}{10(r + 6)(r + 6)} $ $ = \dfrac{ 15(r + 6) + 6(r + 6) - 40} {10(r + 6)(r + 6)} $ Expand: $ = \dfrac{15r + 90 + 6r + 36 - 40}{10r^2 + 120r + 360} $ $ = \dfrac{21r + 86}{10r^2 + 120r + 360}$